The intersection of two functions

Question

How to analyse the intersection of two functions

$y=\ln x+k$ and $y=4x+\ln^4x$

by discussing k.

I tried taking 'ln' on both sides of $\ln x+k - 4x = \ln^4x$ to get rid of the quartic, but it seems didn't work.

That is

$\ln(\ln x+k - 4x )= 4 \ln \ln x$

Let $\psi (x) = \ln(\ln x+k - 4x ) - 4 \ln \ln x$,

such that

$\psi ^{'}(x) = \frac {1/x -4} {\ln x +k -4x} -4/x\ln x$

I still cannot solve this by setting $\psi^{'}(x)=0$.

Any help a hint or a answer will be appreciated. Thank you!

Answer 1

To find the intersection you need to find where, $$ \left(\ln^4 x + 4x\right) - \left(\ln x + k\right)=0 $$

Let $t=\ln x$

$$ t^4 + 4e^t - t - k = 0 $$

which only has solutions in $\mathbb N$ when $k=4,\ t=0 \implies k=4,\ x=1,\ y=4 $.

Answer 2

I think the only solution in N is $$x=1, y=4 and k=4 $$ . One set of solutions in R is $$x=e, y=1+4e$$ and$$k=4e$$. So we may write $$4≤ k ≤ 4e , 1≤ x≤ e$$ and $$4≤ y≤ 1+4e$$ which are the range of intersections.